思路
考虑使用矩阵模拟这个过程。
首先,我们可以设初值为:
\[\begin{bmatrix}
0&1
\end{bmatrix}
\]
表示瘦子初始走 \(0\) 米,胖子初始走 \(1\) 米。
考虑瘦子走一步。
由于瘦子每走一步都不能超过胖子,我们可以使用 \((\min,+)\) 矩乘来维护这个性质。
那么瘦子走一步是:
\[\begin{bmatrix}
1&inf\\
-1&0\\
\end{bmatrix}
\]
同样,胖子走一步是:
\[\begin{bmatrix}
0&inf\\
inf&1\\
\end{bmatrix}
\]
不走是:
\[\begin{bmatrix}
0&inf\\
inf&0\\
\end{bmatrix}
\]
我们可以把每个循环中的所有过程求出来。
由于循环的长度是 \(\operatorname{lcm}(a_1+b_1,a_2+b_2)\)。
我们将相邻的合并以后就只会有最多 \(2(a_1+b_1+a_2+b_2)\) 项。
做一个前缀和即可。
时间复杂度:\(O(n\log n)\)。
可能会有一点卡空间,卡一卡就可以了。
Code
/*! 以渺小启程,以伟大结束。! Created: 2024/07/03 09:01:00
*/
#include <bits/stdc++.h>
using namespace std;#define x first
#define y second
#define int long long
#define mp(x, y) make_pair(x, y)
#define eb(...) emplace_back(__VA_ARGS__)
#define fro(i, x, y) for (int i = (x); i <= (y); i++)
#define pre(i, x, y) for (int i = (x); i >= (y); i--)
inline void JYFILE19();using i64 = long long;
using PII = pair<int, int>;bool ST;
const int N = 1e6 + 10;
const int I = 1e18 + 10;
const int mod = 998244353;struct Mat {int a00, a01, a10, a11;inline friend Mat operator*(const Mat&a, const Mat&b) {return {min(a.a00 + b.a00, a.a01 + b.a10),min(a.a00 + b.a01, a.a01 + b.a11),min(a.a10 + b.a00, a.a11 + b.a10),min(a.a10 + b.a01, a.a11 + b.a11)};}
};
const Mat sz = {1, I,-1, 0};
const Mat pz = {0, I, I, 1};
const Mat sj = {0, I, I, 0};
const Mat pj = {0, I, I, 0};
struct Nod {int a00, a01;inline friend Nod operator*(const Nod&a, const Mat&b) {return {min(a.a00 + b.a00, a.a01 + b.a10),min(a.a00 + b.a01, a.a01 + b.a11)};}
};int q[N];
int t[N * 8];
Mat d[N];
Mat c[N * 8];inline Mat power(Mat x, int y) {Mat res = x; y--;while (y) {if (y & 1) res = res * x;x = x * x, y /= 2;}return res;
}inline void solve() {int a1, b1, a2, b2, n;cin >> a1 >> b1;cin >> a2 >> b2;cin >> n;Nod cs = {0, 1};int p1 = 0, p2 = 0, s1 = 0, s2 = 0, p = 0, ti = 0;Mat zy;do {Mat cur;int sy = 1e9;if (p1 == 0) cur = pz, sy = min(a1 - s1, sy);if (p1 == 1) cur = pj, sy = min(b1 - s1, sy);if (p2 == 0) cur = cur * sz, sy = min(a2 - s2, sy);if (p2 == 1) cur = cur * sj, sy = min(b2 - s2, sy);++p, t[p] = sy, c[p] = cur, ti += t[p];s1 += sy, s2 += sy;if (p1 == 0 && s1 == a1) p1 = 1, s1 = 0;if (p1 == 1 && s1 == b1) p1 = 0, s1 = 0;if (p2 == 0 && s2 == a2) p2 = 1, s2 = 0;if (p2 == 1 && s2 == b2) p2 = 0, s2 = 0;} while (p1 || p2 || s1 || s2);fro(i, 1, p) t[i] = t[i - 1] + t[i];fro(i, 1, n) {cin >> q[i];int w = q[i], l = w / ti;if (l) {w -= ti * l;}if (w) {int x = upper_bound(t + 1, t + p + 1, w) - t;d[i] = c[x];}}fro(i, 1, p) c[i] = power(c[i], t[i] - t[i - 1]);fro(i, 2, p) c[i] = c[i - 1] * c[i];fro(i, 1, n) {Nod ed = cs;int l = q[i] / ti;if (l) {ed = ed * power(c[p], l), q[i] -= ti * l;}if (q[i]) {int x = upper_bound(t + 1, t + p + 1, q[i]) - t;if (x - 1)ed = ed * c[x - 1], q[i] -= t[x - 1];if (q[i])ed = ed * power(d[i], q[i]);}cout << ed.a00 << "\n";}
}signed main() {JYFILE19();int t;cin >> t;while (t--) {solve();}return 0;
}bool ED;
inline void JYFILE19() {// freopen("", "r", stdin);// freopen("", "w", stdout);srand(random_device{}());ios::sync_with_stdio(0), cin.tie(0);double MIB = fabs((&ED - &ST) / 1048576.), LIM = 1024;cerr << "MEMORY: " << MIB << endl, assert(MIB <= LIM);
}